Elegant method to find complex roots easily

We will try to discover the complex roots of the equation f(x)=x²-2x+8.
If we look closely, we see a curve that does not cross the x'Ox axis, and therefore has no real roots.
We notice that the derivative y²=2x-2 finds its cancellation at x=1, and therefore that the point S(1,7) represents a vertex.
Other points can be placed on this curve.
(2,8), (3,11), (4,16), (-1,11), (-2,16), (0,8).
But these are the two complex roots that we want to find.
What are they?
To find out, we must perform a 180° rotation around the point M(0,8) which marks the passage of the curve on the vertical axis y'Oy.
We then obtain the tangent mirror curve at M(0,8) of f(x), which we will call curve g(x), and whose equation will become g(x)=-x²-2x+8.
This curve g(x) has two real reactants x'=-4 and x"=2.
We have said many times that the complex roots of a curve are the real roots of its mirror curve (rotated by 180°) and vice versa.
We therefore have a solution x'=4i and x"=-2i for f(x).
Pay attention to the signs. The point A of g(x) is A(-4,0) which is written A(4i,0) if we describe it via f(x). The point B(2,0) of g(x) is the same for f(x), but is written B(-2i, 0), and not B(2,0).
Here is an elegant method to find complex roots easily, and place them directly on a Cartesian coordinate system.
Numerical application:
f(x)=x²-2x+8.
f(x)=(4i)²-2(4i)+8=16+8+8=0
f(x)=(-2i)²-2(-2i)+8=-4-4+8=0
R.H.